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Eigenvalues and eigenfunctions of Hamiltonian written in terms of ladder operators

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Thunder_Jet
#1
Oct2-11, 12:37 PM
P: 18
Hi everyone!

I am answering this problem which is about the eigenvalues and eigenfunctions of the Hamiltonian given as:

H = 5/3(a+a) + 2/3(a^2 + a+^2), where a and a+ are the ladder operators.

It was given that a = (x + ip)/√2 and a+ = (x - ip)/√2. Furthermore, x and p satisfies the commutation relation [x,p] = i, i.e., p = -i (d/dx).

The question is find the energy eigenvalues and ground state eigenfunction. Is this problem related to the quantum harmonic oscillator? I can't solve it using the usual Hψ = Eψ approach.

Thanks a lot!
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Bill_K
#2
Oct2-11, 12:52 PM
Sci Advisor
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Bill_K's Avatar
P: 4,160
I think you'll see more clearly what to do if you write H entirely in terms of x and p.
jfy4
#3
Oct2-11, 10:03 PM
jfy4's Avatar
P: 647
To find the energy eigenfunction for the ground state, recall that
[tex]
\hat{a}|0\rangle = 0
[/tex]
and recall the definition for [itex]\hat{a}[/itex] in terms of [itex]\hat{x}[/itex] and [itex]\hat{p}[/itex] like Bill said.

Thunder_Jet
#4
Oct2-11, 10:53 PM
P: 18
Eigenvalues and eigenfunctions of Hamiltonian written in terms of ladder operators

Thanks jfy4!

So the problem is really connected to the Harmonic oscillator. Upon simplification of the Hamiltonian, it reads H = 2/3(x^2 + p^2 -1). Substituting p = -i(dx/dt) will result to a 2nd order differential equation. But it is quite difficult to solve the differential equation. Any further suggestions? How would I apply aψ = 0ψ in this case?
jfy4
#5
Oct2-11, 11:03 PM
jfy4's Avatar
P: 647
unless I am mistaken, I don't think that 3 should be in there...

Take
[tex]
\hat{a}|0\rangle =0
[/tex]
and literally write it out in terms of x and p as operators. Then consider
[tex]
\langle x|\hat{a}|0\rangle=0
[/tex]
Also write out [itex]\langle x|0\rangle[/itex] as [itex]\psi_{0}(x)[/itex]. This should get you along.

Forget the 3 comment, I didn't read enough of your post.
Thunder_Jet
#6
Oct2-11, 11:35 PM
P: 18
Yes, that is plausible but I need to use the original Hamiltonian right? Not just the a and a+ operators. Can I use this equation H|ψ> = 0|ψ> for the ground state eigenfunction? The problem is, if I substitute the original Hamiltonian and using H|ψ> = E|ψ>, the differential equation turns out to be something like d^2ψ/dx^2 = 2/3(x^2 + E -1)ψ, which is quite troublesome I think.
jfy4
#7
Oct2-11, 11:48 PM
jfy4's Avatar
P: 647
Perhaps... I just glanced at the OP before posting, and I should have read it a little more carefully... I thought it was the Hamiltonian for the harmonic oscillator, my bad. Now that I work it out though I get a different Hamiltonian than you got. I get
[tex]
H=\frac{3}{2}x^2 +\frac{1}{6}p^2-\frac{5}{6}
[/tex]
Check yours again k.
Thunder_Jet
#8
Oct2-11, 11:57 PM
P: 18
Yes, its possible that I just made some error, but the form is just the same. The problem now would be the differential equation that will arise from the Hamiltonian. Are there any special functions that will solve it or do I need to use methods such as power series, etc, or just introduce a dimensionless quantity. Thanks by the way for your time looking at the problem.
jfy4
#9
Oct3-11, 12:20 AM
jfy4's Avatar
P: 647
Yeah, I suppose you solve this in the coordinate representation...

So get everything away from [itex]\psi ''[/itex] and lump everything in front of [itex]x^2[/itex] and declare a new quantity, q, such that when squared equals that stuff and [itex]x^2[/itex]. Also define a new quantity, [itex]\lambda[/itex], for E and 1 and anything else left over. You should now have a new diffEQ that is a little prettier to solve. Also now let [itex]\psi(x)=u(q)[/itex] and you will have something like
[tex]
u''(q)+(\lambda -q^2)u(q)=0
[/tex]
which is a little more manageable to solve. I have to leave you there sorry, It's late and I have to go to bed. I'll see if I can help more tomorrow.

Good luck, and sorry for not being more help.
Thunder_Jet
#10
Oct3-11, 01:24 AM
P: 18
A, I see. I think it's now a little bit manageable. It is actually similar to the differential equation for the Harmonic oscillator wherein the solutions are Hermite polynomials. Thank you very much for your time! Good night!
Bill_K
#11
Oct3-11, 02:41 PM
Sci Advisor
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Bill_K's Avatar
P: 4,160
Thunder_Jet, I think you're still missing the point of how easy this is. Any Hamiltonian of the form H = αx2 + βp2 is a Harmonic oscillator. You don't need to solve any differential equations. You can immediately write down the solution, just by taking the known ground state of a harmonic oscillator and rescaling x and p.
jfy4
#12
Oct3-11, 05:46 PM
jfy4's Avatar
P: 647
Thanks Bill, that's something I have been missing too. So the ground state can still be the same with whatever scaling needs to be added in also, good to know.
Thunder_Jet
#13
Oct3-11, 09:40 PM
P: 18
Quote Quote by Bill_K View Post
Thunder_Jet, I think you're still missing the point of how easy this is. Any Hamiltonian of the form H = αx2 + βp2 is a Harmonic oscillator. You don't need to solve any differential equations. You can immediately write down the solution, just by taking the known ground state of a harmonic oscillator and rescaling x and p.
I see. So the problem is merely a Harmonic oscillator with shifted or rescaled x and p. Ok, I will try to arrange the eigenvalue equation to see the rescaling with respect to x and p. So the ground state eigenfunction and eigenvalue should be very similar to the original Harmonic oscillator, right? Thanks a lot!


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